Problem: Simplify the following expression: $t = \dfrac{30y + 3}{15y - 27}$ You can assume $y \neq 0$.
Solution: Find the greatest common factor of the numerator and denominator. The numerator can be factored: $30y + 3 = (2\cdot3\cdot5 \cdot y) + (3)$ The denominator can be factored: $15y - 27 = (3\cdot5 \cdot y) - (3\cdot3\cdot3)$ The greatest common factor of all the terms is $3$ Factoring out $3$ gives us: $t = \dfrac{(3)(10y + 1)}{(3)(5y - 9)}$ Dividing both the numerator and denominator by $3$ gives: $t = \dfrac{10y + 1}{5y - 9}$